Question

What is the surface area using integrals of these two lines. Y=10-x^2 and Y=x^2+2 from the bounds x=2 and x=-2.

Answer 1

Check the picture below.

`\bf \displaystyle\int\limits_(-2)^(2)~[\stackrel{\textit{above function}}{(10-x^2)}~~-~~\stackrel{\textit{below function}}{(x^2+2)}]dx\implies \int\limits_(-2)^(2)~(10-x^2-x^2-2)dx \\\\\\ \displaystyle\int\limits_(-2)^(2)~(8-2x^2)dx\implies \left. 8x\cfrac{}{} \right]_(-2)^(2)-\left. \cfrac{2x^3}{3} \right]_(-2)^(2) \\\\\\ ([16]-[-16])~~-~~\left( \left[ \cfrac{16}{3} \right]-\left[ -\cfrac{16}{3} \right]\right)\implies (32)~~-~~\left( \cfrac{32}{3} \right) \\\\\\ \cfrac{64}{3}\implies 21(1)/(3)`